Question

The solution of differential equation $x{\cos }^{2}ydx=y{\cos }^{2}xdy$ is

• A. $x\tan x-y\tan y-\log \left(\frac{\sec x}{\sec y}\right)=c$
• B. $y\tan x-x\tan y-\log (\sec x\cdot \sec y)=c$
• C. $x\tan x-y\tan y+\log (\sec x\cdot \sec y)=c$
• D. None of the above

Answer 1

The correct option is A $x\tan x-y\tan y-\log \left(\frac{\sec x}{\sec y}\right)=c$

$x{\cos }^{2}ydx=y{\cos }^{2}xdy$
$\Rightarrow \frac{x}{{\cos }^{2}x}dx=\frac{y}{{\cos }^{2}y}dy$
$\Rightarrow x{\sec }^{2}xdx=y{\sec }^{2}ydy$
On integration, we get
$\Rightarrow x\tan x-\int 1\cdot \tan xdx$ $=y\tan y-\int 1\cdot \tan ydy$
$\Rightarrow x\tan x-\log \sec x=y\tan y-\log \sec y+c$
$\Rightarrow x\tan x-y\tan y-(\log \sec x-\log \sec y)=c$
$\Rightarrow x\tan x-y\tan y-\log \left(\frac{\sec x}{\sec y}\right)=c$.